Molar Solubility Calculator

How to Calculate Molar Solubility from Ksp

Understanding how to calculate molar solubility is a critical skill for chemistry students and laboratory professionals dealing with saturated solutions. Molar solubility, usually denoted by S, represents the maximum number of moles of a solute that can dissolve in one liter of solvent before a solid precipitate begins to form.

To calculate molar solubility, you must first understand the Solubility Product Constant ($K_{sp}$).

What is the Solubility Product Constant ($K_{sp}$)?

When a slightly soluble ionic compound dissolves in water, it reaches a dynamic equilibrium. For a generic salt that dissociates into cations (positive ions) and anions (negative ions), the equilibrium equation looks like this:

$A_x B_y (s) \rightleftharpoons x A^{m+} (aq) + y B^{n-} (aq)$

The $K_{sp}$ is the equilibrium constant for this reaction. It is calculated by multiplying the molar concentrations of the dissolved ions, each raised to the power of their stoichiometric coefficients. $K_{sp} = [A^{m+}]^x [B^{n-}]^y$

The Math: Deriving Molar Solubility (S)

If the molar solubility of the original salt is $S$ moles per liter, then when it dissolves:

• The concentration of the cation $[A^{m+}]$ will be $x \cdot S$

• The concentration of the anion $[B^{n-}]$ will be $y \cdot S$

By substituting these back into the $K_{sp}$ equation: $K_{sp} = (xS)^x (yS)^y$ $K_{sp} = x^x \cdot y^y \cdot S^{(x+y)}$

To find the Molar Solubility (S), we rearrange the formula algebraically: $S = \left( \frac{K_{sp}}{x^x \cdot y^y} \right)^{\frac{1}{x+y}}$

Step-by-Step Examples

Example 1: Silver Chloride (AgCl) Silver chloride dissociates into one $Ag^+$ ion and one $Cl^-$ ion.

• Cations ($x$) = 1

• Anions ($y$) = 1

• $K_{sp}$ of AgCl = $1.8 \times 10^{-10}$ Calculation: $K_{sp} = (1S)^1(1S)^1 = S^2$ $S = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \text{ mol/L}$

Example 2: Calcium Fluoride ($CaF_2$) Calcium fluoride dissociates into one $Ca^{2+}$ ion and two $F^-$ ions.

• Cations ($x$) = 1

• Anions ($y$) = 2

• $K_{sp}$ of $CaF_2$ = $3.9 \times 10^{-11}$ Calculation: $K_{sp} = (1S)^1(2S)^2 = S \cdot 4S^2 = 4S^3$ $S = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} \approx 2.14 \times 10^{-4} \text{ mol/L}$

Why Use an Online Calculator?

While the formula for $AgCl$ ($S^2$) and $CaF_2$ ($4S^3$) might be easy to memorize, calculating the molar solubility for complex salts like $Ag_3PO_4$ ($27S^4$) or $Bi_2S_3$ ($108S^5$) requires scientific calculators and meticulous algebra.

Our Molar Solubility Calculator simplifies the entire process. By simply entering the $K_{sp}$ in scientific notation and inputting the ratio of your ions, the tool instantly outputs the precise molar solubility in mol/L, as well as the exact equilibrium concentrations of both the cations and anions.